d x sin ( x - a ) sin ( x - b ) is equal to:

Question

$\int \frac{d x}{\sin (x - a) \sin (x - b)}$ is equal to:

Smart Achievers · Accepted Answer

(3)

Ans. $c o s e c (b - a) \log$ $| \frac{\sin (x - b)}{\sin (x - a)} |$ $+ C$

Explanation:

$Let I = \int \frac{d x}{\sin (x - a) \sin (x - b)}$

$= \frac{1}{\sin (b - a)} \int \frac{\sin (b - a)}{\sin (x - a) \sin (x - b)} d x$

$= \frac{1}{\sin (b - a)} \int \frac{\sin (x - a - x + b)}{\sin (x - a) \sin (x - b)} d x$

$= \frac{1}{\sin (b - a)} \int \frac{\sin {(x - a) - (x - b)}}{\sin (x - a) \sin (x - b)} d x$

$= \frac{1}{\sin (b - a)} \times \int \frac{\sin (x - a) \cos (x - b) - \cos (x - a) \sin (x - b)}{\sin (x - a) \sin (x - b)} d x$

$= \frac{1}{\sin (b - a)} \int [\cot (x - b) - \cot (x - a)] d x$

$= \frac{1}{\sin (b - a)} [\log | \sin (x - b) | - \log | \sin (x - a) |] + C$

$= cosec (b - a) \log$ $| \frac{\sin (x - b)}{\sin (x - a)} |$ $+ C$

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